What is the solution for y^{''}-4y^'+4y=5t^2 ?

The solution for y^{''}-4y^'+4y=5t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+(5t^2)/4+(5t)/2+15/8

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y^{''}-4y^'+4y=5t^2

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 t^{2}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 4 y = 5 t^{2}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

y = c_{1} e^{2 t} + c_{2} t e^{2 t} + \frac{5 t ^{2}}{4} + \frac{5 t}{2} + \frac{15}{8}

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} + 64 y = (x + 1) e^{x}

y^{^{''}} - 5 y^{^{'}} + 2 y = x e^{x}

y^{^{''}} + 3 y^{^{'}} - 4 y = 3 e^{x}

y^{^{''}} - y^{^{'}} + y = 12 sin (t) - 4 e^{2 t}

y^{^{''}} - y^{^{'}} - 2 y = 1 + t

What is the solution for y^{''}-4y^'+4y=5t^2 ?
The solution for y^{''}-4y^'+4y=5t^2 is y=c_{1}e^{2t}+c_{2}te^{2t}+(5t^2)/4+(5t)/2+15/8